A more Intuitive proof of regularity of topological group [closed]
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Does someone knows a good reference for the following result?
"A topological group is Hausdorff if and only if the identity is closed."
I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.
general-topology topological-groups
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closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Does someone knows a good reference for the following result?
"A topological group is Hausdorff if and only if the identity is closed."
I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.
general-topology topological-groups
$endgroup$
closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
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– Randall
Dec 12 '18 at 12:17
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Maybe you can list some because I can't even find a single Wikipedia page for it
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– YuiTo Cheng
Dec 12 '18 at 12:21
add a comment |
$begingroup$
Does someone knows a good reference for the following result?
"A topological group is Hausdorff if and only if the identity is closed."
I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.
general-topology topological-groups
$endgroup$
Does someone knows a good reference for the following result?
"A topological group is Hausdorff if and only if the identity is closed."
I have seen proofs in lecture notes of courses on the web, but I would like a reference in a book or an article, in order to refer to it.
general-topology topological-groups
general-topology topological-groups
edited Jan 15 at 9:21
YuiTo Cheng
asked Dec 12 '18 at 11:46
YuiTo ChengYuiTo Cheng
1,325526
1,325526
closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf Dec 13 '18 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MPW, Lord Shark the Unknown, KReiser, Cesareo, supinf
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17
$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21
add a comment |
2
$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17
$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21
2
2
$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17
$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17
$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21
$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21
add a comment |
1 Answer
1
active
oldest
votes
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Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.
Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.
This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.
The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.
Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.
This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.
The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.
$endgroup$
add a comment |
$begingroup$
Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.
Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.
This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.
The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.
$endgroup$
add a comment |
$begingroup$
Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.
Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.
This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.
The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.
$endgroup$
Let $G$ be a topological group. By homogeneity, it is enough to prove that for every open set $V$ containing the neutral element $1$ there is an open set $U$ such that $$1in Usubseteq bar Usubseteq V.$$ Consider the map $mu:Gtimes Gto G$ defined by $$mu(g,h)= gcdot h^{-1}.$$ Since $G$ is a topological group, this map is continuous, and therefore $mu^{-1}(V)$ is open. It also contains the point $(1,1)$, so there is some open set $U$ containing $1$ such that $Utimes Usubseteq mu^{-1}(V)$ (basically by definiton of the product topology). This means that for every $u_1,u_2in U$ we have that $u_1cdot u_2^{-1}in V$ (and in particular $Usubseteq V$ since $1in U$). I claim that this $U$ is the one we are after; since we already have $1in Usubseteq V$, we just need to prove $bar Usubseteq V$.
Let $xin bar U$. The set $$xU={xcdot u|uin U}$$ is an open neighbourhood of $x$ and therefore intersects $U$. That is, there are $u_1,u_2in U$ such that $xcdot u_1 =u_2$. Therefore $x=u_1cdot u_2^{-1}in mu(Utimes U)subseteq V$, as needed.
This is probably the exact same proof as in Munkres only with a different choice of function as to avoid somewhat artificially restricting to symmetric neighbourhoods. To me it seems pretty intuitive.
The main ideas are that $mu$ is continuous and that if $W$ is an open set and $S$ is any set then $Scdot W$ contains in the closure of $S$.
answered Dec 12 '18 at 12:04
CronusCronus
1,098518
1,098518
add a comment |
add a comment |
2
$begingroup$
I don’t understand what’s “ad hoc” about the proof. Symmetric neighborhoods are incredibly useful.
$endgroup$
– Randall
Dec 12 '18 at 12:17
$begingroup$
Maybe you can list some because I can't even find a single Wikipedia page for it
$endgroup$
– YuiTo Cheng
Dec 12 '18 at 12:21